Answer:
The p-value of the test is of 0.2776 > 0.01, which means that the we accept the null hypothesis, that is, the manager's claim that this is only a sample fluctuation and production is not really out of control.
Step-by-step explanation:
A manufacturer considers his production process to be out of control when defects exceed 3%.
At the null hypothesis, we test if the production process is in control, that is, the defective proportion is of 3% or less. So

At the alternate hypothesis, we test if the production process is out of control, that is, the defective proportion exceeds 3%. So

The test statistic is:

In which X is the sample mean,
is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
0.03 is tested at the null hypothesis
This means that 
In a random sample of 100 items, the defect rate is 4%.
This means that 
Value of the test statistic:



P-value of the test
The p-value of the test is the probability of finding a sample proportion above 0.04, which is 1 subtracted by the p-value of z = 0.59.
Looking at the z-table, z = 0.59 has a p-value of 0.7224
1 - 0.7224 = 0.2776
The p-value of the test is of 0.2776 > 0.01, which means that the we accept the null hypothesis, that is, the manager's claim that this is only a sample fluctuation and production is not really out of control.